Question 363964
{{{ 16(y-3)^2+1=8(y-3) }}},
{{{ 16(y-3)^2-8(y-3)+1 = 0 }}},
{{{(4(y-3)-1)^2 = 0}}},
{{{(4y-13)^2 = 0}}},
which means 4y- 13 = 0, or y = 13/4.